Trying to understand a visualization of contravariant and covariant bases

Considering that you're trying to understand a definition, here's a simple example in which you can understand more the definition.

Let's consider 2 vectors, $mathbf{e}_1$ and $mathbf{e}_2$ as they said in the answers (@y255yan) with arbitrary magnitude and directions, and let us consider a vector $mathbf{A}=mathbf{OM}$ represented in this figure

The parallel to the line carrying $mathbf{e}_2$ and passing through $M$ defines another point $M'$, s.t $mathbf{OM'}=x^1mathbf{e}_1$, the same thing with $mathbf{OM''}=x^2mathbf{e}_2$.

Now using the definition: $$mathbf{x}=x^imathbf{e}_i text{Einstein notation}$$

We have the following: $$mathbf{A}=x^1mathbf{e}_1+x^2mathbf{e}_2$$ And here $x^1$ and $x^2$ are the contravariant components od the vector $A$, and to express the covariant components we use the classic definition of scalar product: $$x_1=mathbf{A} . mathbf{e}_1=vertvertmathbf{A}vertvert vertvertmathbf{e}_1vertvert cos alpha \ x_2= mathbf{A} . mathbf{e}_2=vertvertmathbf{A}vertvert vertvertmathbf{e}_2vertvert sin beta $$

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If the vectors $vertvertmathbf{e}_1vertvert=vertvertmathbf{e}_2vertvert=1$ then the orthogonal projections $m'$ and $m''$ of $M$ represent the covariant components of $mathbf{A}$.

I find the convention used here not quite right. I would opt for the opposite. Since the coordinates can be treated as scalar fields $X^a$, their differentials can be written as $dX^a = e^a_{; i} dx^i = vec{e}^a dvec{x}$ (w.r.t. abstract coordinates $x^i$). Analogously, the gradients w.r.t. $X^a$ can be written as $partial_a = e_{a}^{; i} partial_i = vec{e}_a vecnabla$. Hence, the co-frame $vec{e}^a$ is parallel to the coordinate lines whereas the frame $vec{e}_a$ is orthogonal.

It would be more intuitive if you look at the illustrations form Daniel A. Fleisch's book on A Student's Guide to Vectors and Tensors.

Here they are:

The parallel projections represent contravariant components of a given vector $vec{A}$ and the perpendicular projections represent the covariant components of the given vector $vec{A}$.

I'm going to use the definition of a vector from middle school: an object with both direction and magnitude. This definition is enough to resolve your problem. We see by definition, the contravariant basis and covariant basis are just two sets of vectors perpendicular to each other in the sense ${bf e}^{i} cdot {bf e}_j =delta^i_j$. The covariant basis vectors are along the coordinate axis and the contravariant basis vectors are perpendicular to the coordinate axis just like what has been shown in the picture. The terms covariant and contravariant only have meanings when you consider the change of coordinates (lines). For example, if you enlarge the angle between two coordinate lines then the covariant basis vectors will follow what you have done to coordinates lines, that is their angle will also be enlarged , but for the contravariant basis vectors, they will behave in a contra way as you can imagine, their angle will become smaller than before. To the second question, yes you can.

Some preliminary notions first. Let $V$ be a $k$-vector space, and we write $v = v^i e_ i in V$. Here ${e_i}$ with $i=1,dots,dim V$ is a set of vectors, whereas $v^i$ is a set of elements of the underlying field $k$ which makes $v$ a $k$-linear combination of vectors, thus a vector itself in $V$.

We now introduce the dual $V^vee = mathrm{Hom}(V,k)$ vector space, of functionals from elements of $V$ to $k$. We write $b^i$ for this basis, such that $f = f_i b^i in V^vee$ is an element of the dual vector space.

We have by definition $b^i(e_j) = delta^i_j$, since remember it is a functional and so it takes as input an element of the vector space and spits out an element of $k$.

Now, the metric tensor provides a canonical isomorphism. If I have some $f in V^vee$ there exists a unique $v in V$ such that $f(w) = (v,w)$ for all $w in V$, where $(cdot,cdot)$ is the inner product.

So we have that this $v$ associated to $f$ must satisfy,

$$f(w) = f_i b^i(w^j e_j) = f_i w_j b^i(e_j) = f_i w^j delta^i_j = f_i w^i = g_{ji} v^j w^i$$

and so we see that the required vector $v$ is such that $f_i = g_{ji} v^j$, i.e. they are related by raising or lowering indices. The metric provides a canonical isomorphism.

Now if we have a manifold $M$, with a chart (coordinate system) $phi = (x^1,dots,x^n): U to mathbb R^n$, then for a point $p in U subset M$, we have a basis for the tangent space at $p$, defined by

$$frac{partial}{partial x^i}(f) = (partial_i (f circ phi^{-1}))(phi(p)) $$

for $f in C^infty(M)$. (The formula is a little messy: $f$ is composed with the inverse $phi^{-1}$, then we differentiate it and evaluate it at the point $p$ that is mapped to an element of $mathbb R^n$.)

In general covariant and contravariant vectors live in different spaces $V$ and $V^*$, the tangent space and its dual; the latter being the set of linear maps $f:Vto {mathbb R}$. As a consequence plotting these distinct objects on a single diagram is misleading.

That being said, once we are given an inner product $g({bf x},{bf y})={bf x}cdot {bf y}$ we can use it to identify $V$ with $V^*$ by mapping $fin V^*$ to ${bf f}in V$ through $f({bf x}) = g({bf f},{bf x})$. Given a set of basis vectors ${bf e}_i$ and their inner product ${bf e}_icdot {bf e}_j= g_{ij}$ we can identify two sets of components of a vector ${bf x}in V$ by writing $$ {bf x}= x^1 {bf e}_1+ x^2 {bf e_2}+cdots = x^i {bf e}_i $$ or $$ x_i = {bf e}_icdot {bf x}= g_{ij}x^j $$ In some applications (solid state physics for example) it is useful to introduce vectors ${bf e}^{*i} in V$ such that ${bf e}^{*i} cdot {bf e}_j =delta^i_j$ so that $x^{i}= {bf x}cdot {bf e}^{*i}$ and that is probably what the figure means.

In the absence of an inner product there is still a set of basis functions ${bf e}^{*i}$ for $V^*$ such that ${bf e}^{*i}({bf e}_j)= delta^i_j$, but this is an evaluation of a function and not an inner product.